关于使用fs.readdir进行异步目录搜索有什么想法吗?我意识到我们可以引入递归,并调用read目录函数来读取下一个目录,但我有点担心它不是异步的…
什么好主意吗?我已经看了node-walk,它很棒,但它不能像readdir那样只给我数组中的文件。虽然
寻找这样的输出…
['file1.txt', 'file2.txt', 'dir/file3.txt']
关于使用fs.readdir进行异步目录搜索有什么想法吗?我意识到我们可以引入递归,并调用read目录函数来读取下一个目录,但我有点担心它不是异步的…
什么好主意吗?我已经看了node-walk,它很棒,但它不能像readdir那样只给我数组中的文件。虽然
寻找这样的输出…
['file1.txt', 'file2.txt', 'dir/file3.txt']
当前回答
异步
const fs = require('fs')
const path = require('path')
const readdir = (p, done, a = [], i = 0) => fs.readdir(p, (e, d = []) =>
d.map(f => readdir(a[a.push(path.join(p, f)) - 1], () =>
++i == d.length && done(a), a)).length || done(a))
readdir(__dirname, console.log)
Sync
const fs = require('fs')
const path = require('path')
const readdirSync = (p, a = []) => {
if (fs.statSync(p).isDirectory())
fs.readdirSync(p).map(f => readdirSync(a[a.push(path.join(p, f)) - 1], a))
return a
}
console.log(readdirSync(__dirname))
异步读
function readdir (currentPath, done, allFiles = [], i = 0) {
fs.readdir(currentPath, function (e, directoryFiles = []) {
if (!directoryFiles.length)
return done(allFiles)
directoryFiles.map(function (file) {
var joinedPath = path.join(currentPath, file)
allFiles.push(joinedPath)
readdir(joinedPath, function () {
i = i + 1
if (i == directoryFiles.length)
done(allFiles)}
, allFiles)
})
})
}
readdir(__dirname, console.log)
注意:两个版本都将跟随符号链接(与原始fs.readdir相同)
其他回答
香草ES6 +异步/等待+小和可读
我没有在这篇文章中找到我想要的答案;在不同的答案中有一些相似的元素,但我只想要一些简单易读的东西。
为了防止它在未来帮助到任何人(比如几个月后的我自己),这就是我最终使用的:
const { readdir } = require('fs/promises');
const { join } = require('path');
const readdirRecursive = async dir => {
const files = await readdir( dir, { withFileTypes: true } );
const paths = files.map( async file => {
const path = join( dir, file.name );
if ( file.isDirectory() ) return await readdirRecursive( path );
return path;
} );
return ( await Promise.all( paths ) ).flat( Infinity );
}
module.exports = {
readdirRecursive,
}
为了以防有人发现它有用,我还整理了一个同步版本。
var walk = function(dir) {
var results = [];
var list = fs.readdirSync(dir);
list.forEach(function(file) {
file = dir + '/' + file;
var stat = fs.statSync(file);
if (stat && stat.isDirectory()) {
/* Recurse into a subdirectory */
results = results.concat(walk(file));
} else {
/* Is a file */
results.push(file);
}
});
return results;
}
提示:在筛选时使用更少的资源。这个函数本身的过滤器。例如:替换results.push(文件);下面的代码。根据需要调整:
file_type = file.split(".").pop();
file_name = file.split(/(\\|\/)/g).pop();
if (file_type == "json") results.push(file);
短小、现代、高效:
import {readdir} from 'node:fs/promises'
import {join} from 'node:path'
const deepReadDir = async (dirPath) => await Promise.all(
(await readdir(dirPath, {withFileTypes: true})).map(async (dirent) => {
const path = join(dirPath, dirent.name)
return dirent.isDirectory() ? await deepReadDir(path) : path
}),
)
特别感谢函数提示使用{withFileTypes: true}。
这将自动将每个嵌套路径折叠成一个新的嵌套数组。例如,如果:
await deepReadDir('src')
返回如下内容:
[
[
'src/client/api.js',
'src/client/http-constants.js',
'src/client/index.html',
'src/client/index.js',
[ 'src/client/res/favicon.ico' ],
'src/client/storage.js'
],
[ 'src/crypto/keygen.js' ],
'src/discover.js',
[
'src/mutations/createNewMutation.js',
'src/mutations/newAccount.js',
'src/mutations/transferCredit.js',
'src/mutations/updateApp.js'
],
[
'src/server/authentication.js',
'src/server/handlers.js',
'src/server/quick-response.js',
'src/server/server.js',
'src/server/static-resources.js'
],
[ 'src/util/prompt.js', 'src/util/safeWriteFile.js' ],
'src/util.js'
]
但如果你想,你可以很容易地把它压平:
(await deepReadDir('src')).flat(Number.POSITIVE_INFINITY)
[
'src/client/api.js',
'src/client/http-constants.js',
'src/client/index.html',
'src/client/index.js',
'src/client/res/favicon.ico',
'src/client/storage.js',
'src/crypto/keygen.js',
'src/discover.js',
'src/mutations/createNewMutation.js',
'src/mutations/newAccount.js',
'src/mutations/transferCredit.js',
'src/mutations/updateApp.js',
'src/server/authentication.js',
'src/server/handlers.js',
'src/server/quick-response.js',
'src/server/server.js',
'src/server/static-resources.js',
'src/util/prompt.js',
'src/util/safeWriteFile.js',
'src/util.js'
]
TypeScript中基于承诺的递归解决方案,使用Array.flat()处理嵌套返回。
import { resolve } from 'path'
import { Dirent } from 'fs'
import * as fs from 'fs'
function getFiles(root: string): Promise<string[]> {
return fs.promises
.readdir(root, { withFileTypes: true })
.then(dirents => {
const mapToPath = (r: string) => (dirent: Dirent): string => resolve(r, dirent.name)
const directoryPaths = dirents.filter(a => a.isDirectory()).map(mapToPath(root))
const filePaths = dirents.filter(a => a.isFile()).map(mapToPath(root))
return Promise.all<string>([
...directoryPaths.map(a => getFiles(a, include)).flat(),
...filePaths.map(a => Promise.resolve(a))
]).then(a => a.flat())
})
}
这是另一个实现。上述解决方案都没有任何限制,因此如果您的目录结构很大,它们都会崩溃并最终耗尽资源。
var async = require('async');
var fs = require('fs');
var resolve = require('path').resolve;
var scan = function(path, concurrency, callback) {
var list = [];
var walker = async.queue(function(path, callback) {
fs.stat(path, function(err, stats) {
if (err) {
return callback(err);
} else {
if (stats.isDirectory()) {
fs.readdir(path, function(err, files) {
if (err) {
callback(err);
} else {
for (var i = 0; i < files.length; i++) {
walker.push(resolve(path, files[i]));
}
callback();
}
});
} else {
list.push(path);
callback();
}
}
});
}, concurrency);
walker.push(path);
walker.drain = function() {
callback(list);
}
};
使用50的并发工作得非常好,并且几乎和小型目录结构的简单实现一样快。